For a constant unitary matrix $h$, ${\varphi }^{\prime }=h\varphi$ gives ${\varphi }^{\prime †}{\varphi }^{\prime }={\varphi }^{†}{h}^{†}h\varphi ={\varphi }^{†}\varphi$. Also ${\partial }_{\mu }{\varphi }^{\prime }=h{\partial }_{\mu }\varphi$, so the kinetic contraction is unchanged. The mass term and the squared norm in the potential are unchanged for the same reason. The original scalar action has the stated global symmetry. For a spacetime-dependent $h$, however, ${\partial }_{\mu }(h\varphi )=h{\partial }_{\mu }\varphi +({\partial }_{\mu }h)\varphi$, and the extra term prevents local invariance of the ordinary-derivative kinetic term.

Choose the gauge covariant derivative $D_{\mu }={\partial }_{\mu }+ig{A}_{\mu }$, with $A_{\mu }=A_{\mu }^{\alpha }{T}_{\alpha }$ Hermitian. Demand $D_{\mu }^{\prime }(h\varphi )=h{D}_{\mu }\varphi$. Expanding both sides yieldsso the necessary Yang-Mills gauge transformation isUnitarity makes the transformed connection Hermitian. This formula includes the inhomogeneous derivative term; mere conjugation would be insufficient for a local transformation.

Define the Yang-Mills field strength by $[D_{\mu },D_{\nu }]=ig{F}_{\mu \nu }$. Direct expansion givesChoose an orthonormal Hermitian generator basis with $\mathrm{tr}(T_{\alpha }{T}_{\beta })=\kappa {\delta }_{\alpha \beta }$. The compact simple Lie algebra has an invariant positive trace metric, so this basis can be chosen; its structure constants are totally antisymmetric. With these conventions a locally invariant dynamical action isEach matter contraction is invariant because $D\varphi$ transforms like $\varphi$, and the gauge kinetic term is invariant by cyclicity of the trace. It is $-\frac{1}{4}{F}_{\mu \nu }^{\alpha }{F}_{\alpha }^{\mu \nu }$ in components. Choosing $\kappa =1/2$ gives the familiar $-\frac{1}{2}\mathrm{tr}{F}^2$ convention.

Formally introduce a gauge-field source $J_{\alpha }^{\mu }$ and a normalized generating functionalA gauge-field Green function is obtained by source differentiation:Here $\mathcal{N}$ is the unnormalized integral at zero source, so the normalized functional obeys $Z[0]=1$. This expression needs gauge fixing: the integral repeats every physical configuration over its whole gauge orbit, producing an irrelevant but divergent gauge-group volume, and the free gauge kinetic operator has longitudinal zero modes, so it has no inverse propagator on that unrestricted space. Ordinary gauge-field Green functions are gauge-dependent; physical gauge-invariant quantities must be obtained consistently from the gauge-fixed theory.

Use the Lorenz condition ${\chi }^{\alpha }[A]={\partial }^{\mu }{A}_{\mu }^{\alpha }$. To keep signs explicit, parametrize a small transformation by $h=1-ig{\omega }^{\alpha }{T}_{\alpha }+O({\omega }^2)$, equivalently the original parameter is ${\theta }^{\alpha }=g{\omega }^{\alpha }$. The finite transformation above gives the adjoint gauge variation for a positive-sign covariant derivative:Therefore $\delta {\chi }^{\alpha }={\partial }^{\mu }{\mathcal{D}}_{\mu }^{\alpha \beta }{\omega }^{\beta }$. Define the Faddeev-Popov operator as $\mathcal{M}=-{\partial }^{\mu }{\mathcal{D}}_{\mu }$, absorbing the field-independent minus sign of its determinant into normalization.

In a perturbative gauge patch, with residual zero modes removed by boundary conditions, the Faddeev-Popov gauge-orbit identity isIt follows by changing coordinates from the gauge parameter to the gauge condition: the functional Jacobian is precisely the determinant of its linearization. Insert the identity into the integral and change variables along the orbit. The action and measure are invariant, so the gauge-group volume factors out and is divided away. Averaging the gauge-slice value $f$ with Gaussian weight $\exp [-i\int f^2/(2\xi )]$ changes the delta-functional restriction to the covariant gauge-fixing termThe remaining determinant depends on $A$ through ${\mathcal{D}}_{\mu }$. Omitting it while adding only ${\mathcal{L}}_{\mathrm{gf}}$ would weight different orbits incorrectly.

For a finite matrix $M$, expansion of a Grassmann integral selects the product containing every $c$ and $\bar{c}$. Anticommuting the variables into a fixed order attaches the sign of each permutation, giving $\int d\bar{c}dc{e}^{i\bar{c}Mc}$ proportional to $\det M$, with only a matrix-independent phase. The regulated functional version therefore represents the Faddeev-Popov determinant asThe independent Faddeev-Popov ghost fields and antighost fields obey $c_{\alpha }{c}_{\beta }=-c_{\beta }{c}_{\alpha }$, with the corresponding mixed and antighost anticommutation. A commuting complex Gaussian integral instead gives an inverse determinant, so it cannot supply this Jacobian. The required local ghost term isAfter integration by parts,This exhibits the free ghost kinetic term and the ghost-gluon vertex. Ghosts are Grassmann-valued scalar auxiliary fields in the Adjoint representation, not physical fermionic particles or external asymptotic states. Their loop signs arise from anticommutation. In an Abelian theory the adjoint derivative loses the field-dependent part, so the determinant is field-independent and the ghosts decouple; in the non-Abelian theory they must be retained.

The correctly gauge-fixed source functional thus uses $S_{\mathrm{inv}}+S_{\mathrm{gf}}+S_{\mathrm{gh}}$ and integrates over $A,\varphi ,{\varphi }^{†},c,\bar{c}$. Expanding its interaction terms gives the gauge, matter and ghost vertices, and differentiating its source functional computes the gauge-field Green functions. For example, at non-null momentum the quadratic gauge kernel iswhere $P_T$ and $P_L$ are the transverse projector of a vector field and longitudinal projector of a vector field. With the Feynman prescription this gives the gauge-boson propagatorFor the chosen ghost normalization the quadratic kernel is $-{\partial }^2$, so the ghost propagator is $i{\delta }^{\alpha \beta }/(k^2+i0)$. Redefining the antighost sign changes ghost propagator and vertex signs together, without changing the determinant or physical conclusions. Gauge fixing requires the accompanying determinant, and its local representation requires anticommuting ghost and antighost fields. The gauge-orbit argument is local/perturbative: global gauge copies, the Gribov ambiguity, prevent interpreting it as a globally unique gauge slice without further treatment.

Set $ϵ=4-n$ and $a=3/(16{\pi }^2)=-C$. The kinetic term gives the scalar mass dimension $(n-2)/2$, so the quartic bare coupling has mass dimension $4-n=ϵ$. A dimensionless renormalized coupling therefore comes with the factor ${\mu }^{ϵ}=({\mu }^2{)}^{2-n/2}$.

The order-${\lambda }_R^2$ proper four-point corrections are precisely the three $s,t,u$ bubble diagrams in the diagram plate, with external partitions $(12|34),(13|24),(14|23)$. Each carries a symmetry factor one half. Its two-propagator integral has logarithmic ultraviolet divergence in four dimensions, represented in dimensional regularization by a simple pole at $n=4$. That pole's residue is independent of external momenta, so a local quartic counterterm cancels it. The three channels add, rather than describing three separate couplings. The two-point tadpole has no momentum dependence, so it does not supply a one-loop field-normalization correction to the four-point coupling. The first momentum-dependent field correction contributes to this relation only at higher order.

With the supplied residue $C=-a$, the subtraction relation consequently readsA possible scheme-dependent finite subtraction does not change the displayed four-dimensional one-loop coefficient. Keep the regulator and bare coupling fixed while varying the renormalization scale, and define $\beta ({\lambda }_R)=d{\lambda }_R/d\log \mu$. Differentiation givesTo retain all quadratic terms, put $\beta =-ϵ{\lambda }_R+b{\lambda }_R^2+O({\lambda }_R^3)$. The engineering term times the pole factor contributes $-2a{\lambda }_R^2$, and the remaining quadratic coefficient is $a-2a+b$. It vanishes only when $b=a$. ThusDropping the product of the engineering term and pole factor would give the wrong sign for the loop term.

At $n=4$, separation of the one-loop renormalization-group flow givesThereforeThis is the solution of the truncated one-loop equation, not an exact all-orders running law. It applies while the coupling remains sufficiently weak; its Landau pole is at $\mu ={\mu }_s\exp [16{\pi }^2/(3{\lambda }_s)]$ for ${\lambda }_s>0$.

A renormalization-group fixed point has $\beta ({\lambda }^{\ast })=0$. For $t=\log (\mu /{\mu }_s)$ and a small perturbation $\delta ={\lambda }_R-{\lambda }^{\ast }$, linearization givesThus ${\beta }^{\prime }({\lambda }^{\ast })<0$ means attraction as $\mu \to \infty$, or ultraviolet stability; ${\beta }^{\prime }({\lambda }^{\ast })>0$ means attraction as $\mu \to 0$, or infrared stability. If the derivative vanishes, higher terms decide marginal stability. These statements concern the specified coupling direction; other couplings or the mass can be relevant even when this direction is stable.

For $ϵ>0$, the quadratic renormalization-group beta function has zeros $0$ and $ϵ/a$. At the nonzero zero,ConsequentlyThis is the weak-coupling Wilson-Fisher fixed point, perturbatively controlled near four dimensions. Reaching its full critical limit also requires tuning the mass, the thermal relevant direction at the Wilson-Fisher fixed point.

At $n=4$, this nonzero fixed point merges with the Gaussian fixed point. Although the derivative at zero vanishes, $\beta =a{\lambda }_R^2>0$ for a positive weak coupling, so the running solution tends to zero logarithmically toward the infrared. The four-dimensional critical long-distance limit is Gaussian: the quartic interaction is marginally irrelevant, rather than flowing to a nonzero weakly interacting infrared fixed point. This is the Gaussian infrared limit of positive four-dimensional quartic coupling. It is not ultraviolet asymptotic freedom; the positive coupling grows in the ultraviolet. Nor does it say a massive theory has no interactions at finite energy: the critical or massless scaling regime is the one described by this limiting flow. Extrapolating the one-loop law to arbitrarily high cutoff at fixed positive low-energy coupling encounters its Landau pole; avoiding that pole as the cutoff is removed forces the perturbative coupling toward zero. This motivates the usual triviality conclusion, but the one-loop calculation alone is not a nonperturbative theorem excluding all possible strongly coupled ultraviolet completions.

With signature $(+---)$, the momentum-space Feynman rules for phi-fourth theory follow by expanding the interaction exponential and using free-field Wick contractions. Every internal scalar propagator supplies $i/(k^2-m_0^2+i0)$, every quartic Feynman vertex supplies $-i{\lambda }_0$, and each vertex conserves four-momentum. Integrate every independent loop momentum with $d^{4}k/(2\pi {)}^4$ and divide by the Feynman-diagram symmetry factor $S$. Equivalently, assign momenta to all lines and attach $(2\pi {)}^4{\delta }^4(\sum k)$ at each vertex before eliminating constrained integrations. A connected amputated graph therefore has the formwith the remaining overall momentum-conservation delta function understood. For an unamputated correlation function, retain the external propagators; scattering amplitudes instead use LSZ reduction. The factor $4!$ in the interaction cancels the number of ways to attach the four identical fields at a vertex, leaving the stated vertex factor.

For a connected Feynman diagram, $L=N_l-N_v+1$. The superficial degree of divergence is the power under simultaneous large rescaling of all independent loop momenta before considering cancellations or divergent subgraphs. Since each integration contributes four powers and each internal propagator removes two,Counting vertex half-edges gives $4N_v=2N_l+N_e$. Substitution yieldsThe connectedness assumption matters: a disconnected interaction graph with $C$ components instead has the corresponding count $D=4C-N_e$. Tree graphs have no loop integral even when this formal power count is nonnegative.

In four dimensions the possible primitively divergent proper vertices have $N_e=2$ or $4$, with $D=2$ or $0$; odd vertices vanish by the $\varphi \mapsto -\varphi$ symmetry. The divergent local structures are a mass term, a kinetic term and a quartic coupling term. If vacuum diagrams are retained, a vacuum-energy term is also required. Thus one can renormalize using a finite family of operators already allowed by the theory's symmetries: four-dimensional phi-fourth theory is perturbatively renormalizable. The superficial count alone is not an all-orders convergence proof. After divergent subgraphs are subtracted with these counterterms, higher-point graphs have no new overall ultraviolet divergence; a negative overall $D$ does not by itself exclude subdivergences.

The requested lowest-order graphs and the three channels needed for the coupling correction are shown here:

For the two-point function, the free propagator is order one, while its first interacting correction is the tadpole diagram, with $N_v=N_l=L=1$, $N_e=2$, $S=2$ and $D=2$. Its loop integral behaves as $\int d^{4}k/k^2$ and is quadratically divergent. For the connected four-point function, the lowest graph is the order-${\lambda }_0$ contact vertex, with $N_v=1,N_l=L=0$ and formal $D=0$ but no loop divergence. The first loop correction comprises the $s,t,u$ bubble diagrams. Each has $N_v=2,N_l=2,L=1$, $N_e=4$, $S=2$ and $D=0$, matching its logarithmic integral $\int d^{4}k/(k^2{)}^2$.

If the full four-point correlation function rather than its connected part is considered, there is also an order-one contribution: the sum of the three Wick contractions into two free propagators, with pairings $(12)(34),(13)(24),(14)(23)$. Explicitly, $G_0^{(4)}=G_{12}{G}_{34}+G_{13}{G}_{24}+G_{14}{G}_{23}$, with $G_{ij}$ the free scalar propagator between insertions $i,j$. These disconnected pairings are drawn in the bottom row of the plate. They contain no loop integral and are not contributions to the proper four-point vertex.

A six-point amplitude in phi-fourth theory has a one-loop triangle Feynman diagram with two external legs at each of three vertices. It has $N_v=3,N_l=3,L=1$, $N_e=6$, and $D=-2$. For a fixed external-leg partition its loop integral contains three scalar denominators, for exampleAfter Wick rotation, its ultraviolet radial behavior is ${\int }^{\infty }dk{k}^3/k^6={\int }^{\infty }dk/k^3$, which converges. The triangle has no proper loop subgraph, so no ultraviolet subdivergence is hidden in this example. It generates no new six-field counterterm. Nonzero mass avoids a separate infrared issue; with massless propagators exceptional external momenta can produce infrared singularities without changing this ultraviolet conclusion.

The bare mass is a coefficient in the regulated action, whereas the physical mass is the pole position of the full propagator. Write the amputated two-point insertion as $-i\Sigma (p^2)$. Summing its insertions givesAt order ${\lambda }_0$, only the tadpole contributes. Its factor one half follows directly from Wick pairing: of the four fields at the vertex, $4\cdot 3$ choices attach the two labeled external lines and the remaining pair closes the loop, so $12/4!=1/2$. ThusRotate $k^0=i{k}_E^0$ through the Feynman contour. The $i$ from the measure and the $i$ in the propagator, together with the denominator sign, give a positive Euclidean integral. With spherical cutoff $|k_E|<\Lambda$,This is the Wick-rotated cutoff tadpole mass shift, and it is independent of external momentum. Hence to first order,For a positive stable quartic coupling the first-order shift is positive. Keeping the measured pole fixed while changing the cutoff requires adjusting the bare mass. At $m_0=0$ the bracket is understood by its finite limit ${\Lambda }^2$. There is no momentum-dependent wave-function correction from this one-loop tadpole; such corrections begin at higher order.

Use signature $(+---)$ and covariant one-particle state normalization $⟨p^{\prime }|p⟩=(2\pi {)}^{3}2E_p{\delta }^3(\mathbf{p}-{\mathbf{p}}^{\prime })$. The future scattering condition uses the outgoing asymptotic scalar field:As with the incoming condition, this is understood in suitably smeared scattering matrix elements, not as unrestricted pointwise strong convergence of interacting and free local operators.

The free incoming field has expansionThe creation operator defines $|p⟩=a_{\mathrm{in}}^{†}(p)|0⟩$, with $a_{\mathrm{in}}|0⟩=0$, so $⟨p|{\varphi }_{\mathrm{in}}(x)|0⟩=e^{ipx}$. On the other hand, translation invariance givesMatching the stable one-particle state component in the asymptotic condition fixes the constant overlap. Choose its state phase so that it is real and nonnegative. ThenThe same stable particle appears in the outgoing description, with compatible one-particle phases. Thus $Z$ is the squared overlap measured by wave-function renormalization.

Insert a complete set of four-momentum eigenstates between the fields. For $x^0>0$, translation invariance and Hermiticity giveFor $x^0<0$, reversing the time-ordered product gives the same sum with $e^{i{p}_{\alpha }x}$. Therefore, with continuous-state integrals implicit in the completeness sum,Here $\theta$ is the Heaviside step function.

For the usual spectral representation, take a Lorentz-invariant vacuum, a positive-norm Hilbert space and the future-cone spectrum condition. Also take $⟨0|\varphi |0⟩=0$, as for the unbroken real scalar field, or subtract this constant and use the connected correlation function. Otherwise the vacuum intermediate state adds a disconnected constant not represented by the ordinary mass-shell measure below.

Let $W(x)=⟨0|\varphi (x)\varphi (0)|0⟩$ for that centered field, and use the Fourier transform convention $W(x)=\int d^{4}p(2\pi {)}^{-4}{e}^{-ipx}\tilde{W}(p)$. Define the Källén–Lehmann spectral density by the distributional equalityLorentz invariance makes the weight on each future mass shell a function or measure of $p^2$; the spectrum condition gives $p^2\ge 0$. Its nonnegativity follows from the displayed squared matrix elements. ThusApplying time ordering to this expression mass shell by mass shell proves the Källén–Lehmann spectral representation:The factor $i$ is outside ${\Delta }_F$ under the convention in the question.

The stable one-particle state part of completeness is $\int d^3\mathbf{p}[(2\pi {)}^{3}2E_p{]}^{-1}|p⟩⟨p|$. Its overlap is $\sqrt{Z}$, so its contribution to $W$ is $Z$ times the free mass-$m$ two-point function. HenceOther isolated particles as well as multiparticle states can belong to ${\rho }_{\mathrm{rest}}$. The full scalar propagator accordingly has isolated pole $Z/(p^2-m^2+i0)$, making $Z$ its pole residue.

Finally use canonical field normalization: the canonical momentum is $\dot{\varphi }$ and the equal-time canonical commutation relation is $[\dot{\varphi }(t,\mathbf{x}),\varphi (t,\mathbf{y})]=-i{\delta }^3(\mathbf{x}-\mathbf{y})$. The vacuum commutator $C(x)=W(x)-W(-x)$ has the same spectral decomposition. For each free mass shell,independently of $\sigma$. Comparing with the interacting equal-time commutator gives the canonical scalar spectral sum rule:For the particle genuinely interpolated by the stated asymptotic field, $Z>0$; the general nonnegative bound allows a field with zero overlap. The total weight and the upper bound require this canonical normalization and a well-defined distributional equal-time limit. They do not follow from the asymptotic condition alone: rescaling $\varphi$ rescales both $Z$ and the entire spectral measure.

Write $\beta =1/(k_{B}T)$, with $k_B$ the Boltzmann constant. For independent Ising spins, the one-site partition functions on the two sublattices are $z_B=2\cosh \beta (h+g)$ and $z_W=2\cosh \beta (h-g)$. Differentiating each logarithm with respect to its local field givesAt $h=0$, the two sublattices experience opposite fields, so spin reversal combined with sublattice exchange leaves the Hamiltonian invariant and gives $M_W=-M_B$.

For the antiferromagnetic Ising model, take $N$ sites, equal sublattice sizes and $q=2D$ neighbors per site. There are $Nq/2$ bonds counted once. The two-sublattice Ising mean-field approximation expands each bond asand drops only the product of fluctuations. Summing the remaining terms givesThe printed constant omits $N$; for a total Hamiltonian with unnormalized spin sums, this factor is necessary. Applying the independent-spin calculation now givesThe mean-field approximation is favored by large coordination number: averaging many neighbors reduces local relative fluctuations. The size of critical fluctuations still has to satisfy the Ginzburg criterion; high dimension is a reason for validity, not permission to discard every correlation in every regime.

Antisymmetric solutions and the transition. At $h=0$, putting $M_B=-M_W=M$ makes both equations reduce to $M=\tanh \beta (g+qJM)$. A zero-field continuous transition concerns $g=0$ as well: nonzero $g$ explicitly selects the staggered sign and rounds this critical bifurcation. Put $K=\beta qJ$. The graphical intersections of $M$ and $\tanh (KM)$ give only $M=0$ for $K\le 1$. For $K>1$ there are additionally two nonzero intersections $\pm M_{\ast }$: on $M>0$, $\tanh (KM)-M$ is strictly concave, initially increases, and is negative at $M=1$, so there is exactly one positive root. Spin reversal supplies the negative root. The origin becomes unstable in the staggered direction, whereas the nonzero roots have $K(1-M_{\ast }^2)<1$ and are stable. Expanding about the bifurcation givesThus $M_{-}=(M_B-M_W)/2$ turns on continuously, with $k_B{T}_c=qJ$ and order-parameter critical exponent $1/2$.

Uniform solutions. At $g=0$, imposing $M_B=M_W=M$ instead gives $M=\tanh \beta (h-qJM)$. The function $M-\tanh \beta (h-qJM)$ has derivative $1+\beta qJ{\mathrm{sech}}^2\beta (h-qJM)>0$, is negative at $M=-1$ and positive at $M=1$, and consequently has exactly one zero for every positive temperature. At $h=0$ this zero is $M=0$. Antiferromagnetic exchange opposes uniform polarization, so the uniform branch has no ferromagnetic ordering bifurcation. This does not remove the antisymmetric solutions below $T_c$, or exclude transitions of staggered order in the full problem at nonzero uniform field. In particular, uniqueness on the diagonal $M_B=M_W$ is not uniqueness of all solutions.

Free energy and stability. The appropriate off-shell two-sublattice Ising variational free energy is the energy minus temperature times the independent-spin entropy. DefineThere are $Nq/2$ bonds and $N/2$ spins of each kind, so the free-energy density per site isSince $s^{\prime }(m)=\mathrm{atanh}m$, stationarity givesand the analogous equation for $M_W$; these are exactly the two mean-field self-consistency equations. On a stationary solution, the identity $s(m)=m\mathrm{atanh}m-\ln (2\cosh (\mathrm{atanh}m))$ gives the equivalent on-shell expressionThe entropy form must be used to compare off-shell variations; treating this latter auxiliary expression as the variational potential would give spurious stability conclusions.

Using $s(m)=-\ln 2+m^2/2+m^4/12+O(m^6)$ and $M_B=M_{+}+M_{-}$, $M_W=M_{+}-M_{-}$ givesThis includes the requested quadratic expansion and first-order field terms. The positive quartic coefficient verifies a continuous phase transition in $M_{-}$ when the staggered quadratic coefficient changes sign at $k_{B}T=qJ$. The uniform quadratic coefficient remains positive.

Susceptibilities and a printed error. On a stable zero-field ordered branch let $M_B=-M_W=M$ and $a=\beta (1-M^2)$. Since the derivative of $\tanh x$ is $1-{\tanh }^{2}x$, differentiating the two equations with respect to the uniform field givesTheir sum and difference yieldThe difference equation is $(1-qJa){\chi }_{-}=0$; stability implies $qJa<1$ for $T<T_c$. Symmetry also explains the zero result: a uniform perturbation changes the two sublattice magnetizations equally to first order, and hence does not linearly change the staggered magnetization. This ${\chi }_{-}$ is a cross response to $h$, not the magnetic susceptibility to the staggered field $g$.

The displayed ${\chi }_{+}$ in the source instead contains $1+M^2$. It cannot follow from differentiating the stated equations. A decisive limiting check is $T\to 0$: $1-M^2$ vanishes exponentially and the actual uniform response tends to zero, whereas the printed expression tends to $1/(qJ)$. Above $T_c$, $M=0$ and the same linear system givesFor comparison, the actual staggered susceptibility is $\partial M_{-}/\partial g=a/(1-qJa)$, becoming $1/(k_{B}T-qJ)$ above $T_c$. Its divergence, rather than a divergence of the uniform response, detects antiferromagnetic criticality.

Use a finite-dimensional complex semisimple Lie algebra, or the complexification of its compact real form, as required for the root construction. Choose a Cartan subalgebra with basis $H_1,\dots ,H_r$. Simultaneous diagonalization of its adjoint action gives the root-space decompositionThe nonzero weights are the roots of a root system; each root space is one-dimensional in this setting, and roots occur in opposite pairs. This supplies the Cartan and root-vector basis.

The Jacobi identity givesIf the bracket is nonzero and $\alpha +\beta \ne 0$, it is a nonzero vector in the root space of weight $\alpha +\beta$, so that sum is a root. The printed implication needs the opposite-root exception: for $\beta =-\alpha$, $[E_{\alpha },E_{-\alpha }]$ can be a nonzero Cartan element, while zero is not a root. The supplied simple-root commutator itself gives this counterexample. This is the root bracket and the opposite-root exception.

Choose a real linear functional nonzero on every root; positive roots are those on which it is positive, and negative roots are their negatives. The simple roots are positive roots not decomposable into two positive roots. They form a basis ${\alpha }_1,\dots ,{\alpha }_r$ of the root span, and every root is an integer combinationwith all $m_i\ge 0$ for a positive root and all $m_i\le 0$ for a negative root. Positive roots can be decomposed repeatedly until simple roots remain; the finite root set makes this procedure terminate.

For $i\ne j$, ${\alpha }_j-{\alpha }_i$ has simple-root coefficients of both signs and is not a root. It is also nonzero, so the root-space bracket rule givesThe operators $E_i^{\pm },{\hat{H}}_i$ have the SU(2) Lie algebra normalization ${\hat{H}}_i=2J_3$, $E_i^{\pm }=J_{\pm }$. In a finite-dimensional representation, a vector killed by $E_i^{-}$ is a lowest-weight vector in its SU(2) summands. If its ${\hat{H}}_i$ value is $-\lambda$, those summands have spin $\lambda /2$, soOne can also read the termination from successive raising coefficients $\sqrt{(\lambda -k)(k+1)}$ starting at the lowest weight.

Apply this result in the finite-dimensional Adjoint representation, taking the lowest-weight vector to be $E_j^{+}$. The vanishing bracket just proved is its lowering condition. If $[{\hat{H}}_i,E_j^{+}]=a_{ij}{E}_j^{+}$, the lowest-weight result forces $a_{ij}=-n_{ij}$ with $n_{ij}$ a nonnegative integer. It also yieldsThe last identity is exactly the requested nested commutator. This is the SU(2) proof of the Serre relations, and the identities are the Serre relations in these Chevalley basis conventions.

To construct a highest-weight representation, start from its highest-weight vector and apply negative-root operators. Fix an order on the negative roots. The Poincaré-Birkhoff-Witt theorem gives spanning vectorsof weight $w-{\sum }_i{k}_i{\beta }_i$. The spanning assertion follows by commuting all negative-root operators to the left, Cartan operators to the middle and positive-root operators to the right; commutators supply lower-order words, positive operators kill the highest-weight vector and Cartan operators act there by scalars. In the finite-dimensional highest-weight representation, impose the simple-root string relations $(E_i^{-}{)}^{w_i+1}|w⟩=0$ and the commutator/Serre relations. At each weight retain a linearly independent set of the resulting words, discarding zero words and dependencies. This gives a basis, with separate independent vectors retained when a weight has multiplicity greater than one. The construction is finite in the assumed finite-dimensional representation; treating all lowering words as automatically independent would overcount.

For SU(3), put $e_i=E_i^{+}$, $f_i=E_i^{-}$ and $h_i={\hat{H}}_i$. The off-diagonal Cartan entries are minus one, so the Serre relations give $(\mathrm{ad}{e}_1{)}^2{e}_2=(\mathrm{ad}{e}_2{)}^2{e}_1=0$. With $e_3=[e_1,e_2]$, these implyFor $f_3=-[f_1,f_2]$, expand the commutator using Jacobi identity and the distinct-root brackets:since $[h_1,f_2]=f_2$. Similarly,and the same calculation on $f_3$ gives $[h_1,f_3]=-f_3$. ThusThese signs also agree with the explicit model $e_1=E_{12}$, $e_2=E_{23}$, $e_3=E_{13}$, $f_1=E_{21}$, $f_2=E_{32}$, $f_3=E_{31}$.